Going from signed integers to unsigned integers and vice versa in C++20

Question

Before C++20 signed integers were not guaranteed to be two's complement. Now we have two papers proposing standardization of two's complement as the only representation: p0907 and p1236 and, if I understand correctly, one of them got merged into C++20 working draft.

So, what does it mean for signed-to-unsigned conversion and vice versa? I've looked at cppreference and has found the following wording:

If the destination type is unsigned, the resulting value is the smallest unsigned value equal to the source value modulo 2n where n is the number of bits used to represent the destination type.

If the destination type is signed, the value does not change if the source integer can be represented in the destination type. Otherwise the result is the unique value of the destination type equal to the source value modulo 2n where n is the number of bits used to represent the destination type. (Note that this is different from signed integer arithmetic overflow, which is undefined).

Unfortunately, I have problems understanding this wording and I want to know what is written in C++20 Working Draft.

So there are two questions:

1. Language lawyer part: can someone please point what the standard says exactly and where in the standard does it say it?

2. Can someone explain that wording in more layman's terms, probably also explaining modular arithmetic and providing examples?

Answer 1

The conversion rule is [conv.integral]/3:

Otherwise, the result is the unique value of the destination type that is congruent to the source integer modulo 2N, where N is the range exponent of the destination type.

where the range exponents are described in table in [basic.fundamental], but mean what you expect (for int it's at least 16, for long long, it's at least 64, etc.).

For instance, converting a short with value -3 to unsigned short is to find the unique value of type unsigned short which is congruent to -3 modulo 216... which is to say, 216-3 or 65533. But converting that same short value of -3 to unsigned long long would change the modulo base to 264, so you end up with 18446744073709551613 instead.

Converting from a signed type to a signed type with a different range exponent (or likewise unsigned to unsigned) is more straightforward - you either just chop off bits or zero-extend them. Converting a short with value 258 to int or long or long long is just 258, but to signed char is 2, for instance.

Answer 2

Your quote uses a lot of words to say something pretty simple: For every integer i, there is exactly one integer k between 0 and M so that i % M == k (in the mathematical sense, not the "fixed size integer representations" sense). In more layman terms, i % M == k means "if I add or subtract M the right number of times from i, I can obtain k".

In the case of integral conversions, M = 2^N where N is the number of bits in the destination type. The standard says:

[conv.integral]#3

Otherwise, the result is the unique value of the destination type that is congruent to the source integer modulo 2^N, where N is the range exponent of the destination type.

In examples:

Say your destination type has 4 bits, so it can represent the 2^4 = 16 values from 0 to 15. Converting 1 into this range yields 1, 14 yields 14, 15 yields 15, 16 yields 0, 17 yields 1, 18 yields 2 and so on. Converting 0 yields 0, -1 yields 15, -2 yields 14 etc.

If you really want a deeper introduction to modular arithmetic, that is beyond the scope of this site. You should refer to the plentiful resources in the web, such as this one.